Q.If $A = \begin{bmatrix}2 & 1\end{bmatrix}$, $B = \begin{bmatrix}5 & 3 & 4\\ 8 & 7 & 6\end{bmatrix}$ and $C = \begin{bmatrix}-1 & 2 & 1\\ 1 & 0 & 2\end{bmatrix}$, verify that $A(B + C) = (AB + AC)$.

Compute both sides of $A(B+C)=AB+AC$. Here $A$ is $1\times 2$, $B,C$ are $2\times 3$.

Left side. $B+C=\begin{bmatrix} 4 & 5 & 5 \\ 9 & 7 & 8 \end{bmatrix}$, so

$A(B+C)=\begin{bmatrix} 2 & 1 \end{bmatrix}\begin{bmatrix} 4 & 5 & 5 \\ 9 & 7 & 8 \end{bmatrix}=\begin{bmatrix} 17 & 17 & 18 \end{bmatrix}.$ …

Sizes first

$A$ is $1\times 2$, and $B,C$ are $2\times 3$. So $B+C$ is $2\times 3$, and every product below is $1\times 3$ — the two sides can indeed be compared.

Left side: $A(B+C)$

Add inside the parentheses first:

Then

Right side: $AB+AC$

$AB=\begin{bmatrix} 2 & 1 \end{bmatrix}\begin{bmatrix} 5 & 3 & 4 \\ 8 & 7 & 6 \end{bmatrix}=\begin{bmatrix} 2(5)+1(8) & 2(3)+1(7) & 2(4)+1(6) \end{bmatrix}=\begin{bmatrix} 18 & 13 & 14 \end{bmatrix},$ …

Method: Verifying left-distributivity $A(B+C)=AB+AC$

Use this to check the distributive law over addition.

Steps

Step 1: Confirm conformability.

$A(B+C)$ needs $B$ and $C$ to share an order that can be left-multiplied by $A$; check the inner dimensions.

Step 2: Compute the left side.

Add $B+C$ entrywise, then multiply $A(B+C)$. …

Common Mistakes

Mistake 1: Distributing on the wrong side.

Why it's wrong: the identity is $A(B+C)=AB+AC$ with $A$ on the left; $(B+C)A$ is a different expression. Correct approach: keep $A$ as the left factor in every product.

Mistake 2: Adding $B$ and $C$ incorrectly.

Why it's wrong: $B+C$ is entrywise ($\begin{bmatrix}4&5&5\\9&7&8\end{bmatrix}$); mismatching positions corrupts the left side. Correct approach: add corresponding entries. …